WAEC Mathematics 2015 Paper

Concept:A perpendicular from the centre of a circle to a chord bisects the chord. The radius, half the chord, and the perpendicular distance form a right-angled triangle.Explanation:Given chord length $AB = 16\sqrt{3}\ \text{cm}$ and perpendicular distance from centre $O$ to chord $OM = 10\ \text{cm}$.Since $OM$ bisects $AB$, we have:$AM = \frac{AB}{2} = \frac{16\sqrt{3}}{2} = 8\sqrt{3}\ \text{cm}$.In right-angled triangle $AMO$, using Pythagoras theorem:$r^2 = AM^2 + OM^2$$r^2 = (8\sqrt{3})^2 + 10^2$$r^2 = 64 \times 3 + 100$$r^2 = 192 + 100 = 292$$r = \sqrt{292} = 17.088\ \text{cm}$Correct to the nearest cm, $r = 17\ \text{cm}$.Answer:$17\ \text{cm}$ (Option B).